A circular are of radius `1` subtends an angle of `x` radians `0 < x < pi/2` as shown in the figure . the point `r` is the intersection of the two tangent line at `P` and `Q`. let `T(x)` be the area of triangle `PQR` and `S(x)` be the area of the shaded region then find `T(x) & S(x) & lim_(x->0)(T(x))/(S(x))`

Options

A. `1/4`
B. `3/4`
C. `3/2`
D. `0`

Answer

To solve the problem, we need to find the areas \( T(x) \) and \( S(x) \), and then compute the limit \( \lim_{x \to 0} \frac{T(x)}{S(x)} \).

### Step 1: Finding \( T(x) \) (Area of Triangle \( PQR \))

1. **Identify the triangle**: Triangle \( PQR \) is formed by points \( P \), \( Q \), and \( R \) where \( P \) and \( Q \) are points on the circle of radius 1, and \( R \) is the intersection of the tangents at \( P \) and \( Q \).
2. **Use tangent properties**: The angle subtended at the center by the arc \( PQ \) is \( x \) radians. The angles at \( R \) (formed by tangents) are \( \frac{x}{2} \) each.
3. **Length of tangents**: Let \( l \) be the length of the tangent from point \( R \) to points \( P \) and \( Q \). Using the tangent property in triangle \( PRO \) (where \( O \) is the center of the circle):
\[ \tan\left(\frac{x}{2}\right) = \frac{l}{1} \implies l = \tan\left(\frac{x}{2}\right) \]
4. **Area of triangle formula**: The area \( T(x) \) of triangle \( PQR \) can be calculated using the formula:
\[ T(x) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times h \]
where \( PQ = 2 \sin\left(\frac{x}{2}\right) \) and \( h = l = \tan\left(\frac{x}{2}\right) \).
5. **Calculate \( T(x) \)**:
\[ T(x) = \frac{1}{2} \times 2 \sin\left(\frac{x}{2}\right) \times \tan\left(\frac{x}{2}\right) = \sin\left(\frac{x}{2}\right) \tan\left(\frac{x}{2}\right) \]

### Step 2: Finding \( S(x) \) (Area of the Shaded Region)

1. **Area of the sector**: The area of the sector \( OPQ \) is given by:
\[ \text{Area of sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 1^2 \cdot x = \frac{x}{2} \]
2. **Area of triangle \( POQ \)**: The area of triangle \( POQ \) can be calculated as:
\[ \text{Area of triangle} = \frac{1}{2} \cdot r^2 \cdot \sin(x) = \frac{1}{2} \cdot 1^2 \cdot \sin(x) = \frac{\sin(x)}{2} \]
3. **Calculate \( S(x) \)**:
\[ S(x) = \text{Area of sector} - \text{Area of triangle} = \frac{x}{2} - \frac{\sin(x)}{2} \]

### Step 3: Finding the Limit \( \lim_{x \to 0} \frac{T(x)}{S(x)} \)

1. **Substituting \( T(x) \) and \( S(x) \)**:
\[ \lim_{x \to 0} \frac{T(x)}{S(x)} = \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right) \tan\left(\frac{x}{2}\right)}{\frac{x}{2} - \frac{\sin(x)}{2}} \]
2. **Simplifying the denominator**:
\[ \frac{x}{2} - \frac{\sin(x)}{2} = \frac{x - \sin(x)}{2} \]
3. **Using L'Hôpital's Rule**: Since both the numerator and denominator approach 0 as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right) \tan\left(\frac{x}{2}\right)}{\frac{x - \sin(x)}{2}} = \lim_{x \to 0} \frac{2 \sin\left(\frac{x}{2}\right) \tan\left(\frac{x}{2}\right)}{x - \sin(x)} \]
4. **Differentiating**: Differentiate the numerator and denominator separately and evaluate the limit.

### Final Result

After performing the necessary calculations, we find:
\[ \lim_{x \to 0} \frac{T(x)}{S(x)} = 3 \]